Datasheet

L1 =
V
OUT
x (V
IN
- V
OUT
)
I
OR
x F
s
x V
IN
F
S(max)2
=
V
IN(min)
- V
OUT
V
IN(min)
x 260 ns
F
S(max)1
=
V
OUT
V
IN(max)
x 200 ns
LM5006
www.ti.com SNVS646B –FEBRUARY 2011–REVISED MARCH 2013
Applications Information
SELECTION OF EXTERNAL COMPONENTS
A guide for determining the component values is illustrated with a design example. Refer to the Block Diagram.
The following steps will configure the LM5006 for:
• Input voltage range (Vin): 15V to 75V
• Output voltage (V
OUT
): 10V
• Load current (for continuous conduction mode): 100 mA to 400 mA
• Switching Frequency: 300 kHz
R
FB1
, R
FB2
: V
OUT
= V
FB
x (R
FB1
+ R
FB2
) / R
FB1
and since V
FB
= 2.5V, the ratio of R
FB2
to R
FB1
calculates as 3:1.
Standard values of 3.01 kΩ and 1.00 kΩ are chosen. Other values could be used as long as the 3:1 ratio is
maintained.
F
s
and R
T
: Unless the application requires a specific frequency, the choice of frequency is generally a
compromise. A higher frequency allows for a smaller inductor, input capacitor, and output capacitor (both in value
and physical size), while providing a lower conversion efficiency. A lower frequency provides higher efficiency,
but generally requires higher values for the inductor, input capacitor and output capacitor. The maximum allowed
switching frequency for the LM5006 is limited by the minimum on-time (200 ns) at the maximum input voltage,
and by the minimum off-time (260 ns) at the minimum input voltage. The maximum frequency limit for each
application is defined by the following two calculations:
(7)
(8)
The maximum allowed frequency is the lesser of the two above calculations. See the graph " Maximum
Switching Frequency". For this exercise, F
s(max)1
calculates to 667 kHz, and F
s(max)2
calculates to 1.28 MHz.
Therefore the maximum allowed frequency for this example is 667 kHz, which is greater than the 300 kHz
specified for this design. Using Equation 2, RT calculates to 258 kΩ. A standard value 261 kΩ resistor is used.
The minimum on-time calculates to 469 ns, and the maximum on-time calculates to 2.28 µs.
L1: The main parameter affected by the inductor is the output current ripple amplitude. The choice of inductor
value therefore depends on both the minimum and maximum load currents, keeping in mind that the maximum
ripple current occurs at maximum Vin.
a) Minimum load current: To maintain continuous conduction at minimum Io (100 mA) if a flyback diode is
used, the ripple amplitude (IOR) must be less than 200 mA p-p so the lower peak of the waveform does not
reach zero. L1 is calculated using the following equation:
(9)
At Vin = 75V, L1(min) calculates to 144µH. The next larger standard value (150 µH) is chosen and with this value
I
OR
calculates to 193 mA p-p at Vin = 75V, and 74 mA p-p at Vin = 15V
b) Maximum load current: At a load current of 400 mA, the peak of the ripple waveform must not reach the
minimum value of the LM5006’s current limit threshold (700 mA). Therefore the ripple amplitude must be less
than 600 mA p-p, which is already satisfied in the above calculation. With L1 = 150 µH, at maximum Vin and Io,
the peak of the ripple is 498 mA. While L1 must carry this peak current without saturating or exceeding its
temperature rating, it also must be capable of carrying the maximum value of the LM5006’s current limit
threshold without saturating, since the current limit is reached during startup.
The DC resistance of the inductor should be as low as possible. For example, if the inductor’s DCR is 0.5 ohm,
the power dissipated at maximum load current is 0.08W. While small, it is not insignificant compared to the load
power of 4W.
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